Question

A person uses a mixture having 8parts pure wine and 3 parts water. After taking some portion of the mixture, he adds equal amount of pure water to the remaining part of mixture such that the amount of alcohol and water is in 2:1 ratio. What part of mixture is taken out?

Answer 1

Given :-

• initial wine : water = 8 : 3
• After taking some portion of the mixture, he adds equal amount of pure water to the remaining part of mixture , wine : water = 2 : 1

To Find :-

• What part of mixture is taken out ?

Solution :-

final concentration formula says that :-

• final concentration which remains same = initial concentration which remains same * (1 - x/V) .
• x = Quantity removed from mixture and quantity added.
• V = Total volume

Given that, some mixture was taken out and adds equal amount of pure water . So, quantity of wine remains same in the mixture as before.

Than,

→ in final mixture = wine : water = 2 : 1

→ in final mixture wine = (2/3) = FC

and,

→ in initial mixture = wine : water = 8 : 3

→ in initial mixture wine = (8/11) = IC

Let :-

• x = part which was removed .
• V = Total quantity .

Putting all values now in formula , we get ,

→ (2/3) = (8/11)(1 - x/V)

→ (1/3) = (4/11)(1 - x/V)

→ (1/3) * (11/4) = 1 - x/V

→ (11/12) = 1 - x/V

→ x/V = 1 - (11/12)

→ x/V = (1/12) (Ans.)

Hence, (1/12) part of Mixture was taken out.